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Question Number 100806 by I want to learn more last updated on 28/Jun/20

Commented by I want to learn more last updated on 28/Jun/20

$33 sir .$
Commented by john santu last updated on 28/Jun/20

$\tan^{2} (\frac{2 π}{9}) + \tan^{2} (\frac{4 π}{9}) + \tan^{2} (\frac{8 π}{9}) = 33$
Commented by I want to learn more last updated on 28/Jun/20

$Workings sir . Please$
	Answered by MJS last updated on 28/Jun/20
	$put tan^2 x =t ⇒ t+((4t)/((t−1)^2 ))+((16t(t−1)^2 )/((t^2 −6t+1)^2 ))=33 ⇒ t^7 −47t^6 +545t^5 −2291t^4 +3611t^3 −2205t^2 +483t−33=0 (t^4 −14t^3 +56t^2 −62t+11)(t^3 −33t^2 +27t−3)=0 now solve this approximately or exactly, for the first factor the exact solution is not useable. for 0≤x<(π/3) it leads to { ((x_1 =π/9)),((x_2 =2π/9)),((x_3 =4π/9)) :} from the 3^(rd) −degree−polynome { ((x_4 ≈.436880)),((x_5 ≈.873645)),((x_6 ≈1.13059)),((x_7 ≈1.22795)) :} from tbe 4^(th) −degree−polynome the solutions of the given equation are x_i +2nπ and (2n+1)π−x_i with 1≤i≤7 and n∈Z$
	$put \tan^{2} x = t$ $\Rightarrow$ $t + \frac{4 t}{{(t - 1)}^{2}} + \frac{16 t {(t - 1)}^{2}}{{(t^{2} - 6 t + 1)}^{2}} = 33$ $\Rightarrow$ $t^{7} - 47 t^{6} + 545 t^{5} - 2291 t^{4} + 3611 t^{3} - 2205 t^{2} + 483 t - 33 = 0$ $(t^{4} - 14 t^{3} + 56 t^{2} - 62 t + 11) (t^{3} - 33 t^{2} + 27 t - 3) = 0$ $now solve this approximately or exactly,$ $for the first factor the exact solution is not$ $useable .$ $for 0 ⩽ x < \frac{π}{3} it leads to$ ${\begin{cases} x_{1} = π / 9 \\ x_{2} = 2 π / 9 \\ x_{3} = 4 π / 9 \end{cases} from the 3^{rd} - degree - polynome$ ${\begin{cases} x_{4} \approx . 436880 \\ x_{5} \approx . 873645 \\ x_{6} \approx 1.13059 \\ x_{7} \approx 1.22795 \end{cases} from tbe 4^{th} - degree - polynome$ $the solutions of the given equation are$ $x_{i} + 2 n π and (2 n + 1) π - x_{i} with 1 ⩽ i ⩽ 7 and n \in Z$
	Commented by maths mind last updated on 28/Jun/20

	$n i c e w o r c k s i r$
	Commented by I want to learn more last updated on 28/Jun/20

	$Thanks sir$
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